The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th
e flow. The differential height between the water columns connected to the two outlets of the probe is 0.126 m.Take the density of water to be 1000 kg/m3. The gas constant of air is R = 0.287 kPa-m3/kg-K.The air temperature and pressure in the duct are 352 K and 98 kPa, respectively.
1 answer:
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
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Consider checking attachment for the step by step solution.
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C
Answer:
The dude was killed around 6:30PM
Explanation:
Newton's law of cooling states:
where,
= initial temp
= temp of room
= temp after t hours
= how fast the temp is changing
= time (hours)
because the body was initlally 31ºC when the police found it
because that was the room temp
because the body temp drop to 30ºC after 1 hour
t = 1 because that's the time it took for the body temp to drop to 30ºC
we don't know k so we must solve for this
rearrange the equation to solve for k
plug in the numbers to solve for k
Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.
because the body was 37ºC right after being killed
because that was the room temp
because the body temp when the police found it
we solved this earlier
t = ??? we don't know how long it took from the time of the murder to when the police found the body
Rearrange the equation to solve for t
plug in the values
t ≈ 4.337 hours from the time the body was killed to when the police found it.
The police found the body at 11:00PM so subtract 4.337 from that.
11 - 4.33 = 6.66 ≈ 6:30PM